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vesna_86 [32]
3 years ago
7

Express the given situation as a linear inequality. needs at least units of a nutritional supplement per day. Red pills provide

units and blue pills provide. Let x be the number of red pills and y be the number of blue pills.
A. 6x+ 5y ≥ 32
B. 11(x + y) ≥ 32
C. 320X+ y) ≥ 11
D. x+y≥32
Mathematics
1 answer:
kakasveta [241]3 years ago
8 0
Wiener eejnejene jenejeeb ewj



Jenejeneb Jenner be.



Hope the helps
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Step-by-step explanation: I did this in school and got it right.

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A certain triangle has two 45° angles. What type of triangle is it?
neonofarm [45]
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4 years ago
Please look at the picture and answer it.
Zanzabum

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6 0
2 years ago
The sum of two numbers is 75. The second number is<br> less than twice the first. Find the numbers.
Lera25 [3.4K]

Answer:

First number =26

Second number =49

Step-by-step explanation:

Question:

The sum of two numbers is 75. The second number is 3 less than twice the first. Find the numbers.

Given:

Sum of two numbers = 75

The second number is 3 less than twice the first.

To find the numbers.

Solution:

Let the first number be = x

Let second number be = y

The sum equation can be given as:

x+y=75

It is given that the second number is 3 less than twice the first.

This can be given as:

y=2x-3

Substituting the value of y in terms of x in the sum equation.

x+(2x-3)=75

Combining like terms.

3x-3=75

Adding 3 both sides.

3x-3+3=75+3

3x=78

Dividing both ides by 3.

\frac{3x}{3}=\frac{78}{3}

x=26

Substitution x=26 in the second equation to find y.

y=2(26)-3\\y=52-3\\\therefore y=49

Thus, first number =26 and second number =49.

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B%20%20%5Csqrt%7B3x%7D%20%7D%7B1%20%2B%20e%5E%7B2x%7D%20%7D%20" id="TexF
kumpel [21]

~~~~~~~~y = \dfrac{\sqrt{3x}}{ 1+ e^{2x}}\\\\\\\implies \dfrac{dy}{dx} = \dfrac{d}{dx}  \left( \dfrac{\sqrt{3x}}{1+ e^{2x}} \right)\\\\\\~~~~~~~~~~~~=\dfrac{(1 + e^{2x} ) \dfrac{d}{dx} \left(\sqrt{3x} \right) - \left(\sqrt{3x} \right)\dfrac{d}{dx}(1+e^{2x})}{\left( 1+ e^{2x} \right)^2}~~~~~~~~~~~~;[\text{Quotient rule}]\\\\\\~~~~~~~~~~~~=\dfrac{(1+e^{2x}) \cdot \dfrac{1}{2\sqrt{3x}} \cdot 3-\left(\sqrt{3x} \right) \cdot 2 e^{2x}}{(1 + e^{2x})^2}~~~~~~~~~~~~~~~~~~~~;[\text{Chain rule}]\\\\\\

            =\dfrac{\dfrac{\sqrt 3 (1 + e^{2x})}{2\sqrt x} -2e^{2x} \sqrt{3x}}{(1+e^{2x})^2}\\\\\\=\dfrac{\tfrac{\sqrt 3(1 +e^{2x}) - 4x\sqrt 3 e^{2x}}{2\sqrt x}}{(1+e^{2x})^2}\\\\\\=\dfrac{\sqrt 3(1+e^{2x} -4xe^{2x})}{2\sqrt x(1 +e^{2x})^2}

7 0
2 years ago
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